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Exercise 16 - Let $(X,M,\mu)$ be a measure space. A set $E\subset X$ is called locally measurable if $E\cap A\in M$ for all $A\in M$ such that $\mu(A) < \infty$. Let $\tilde{M}$ be the collection of all locally measurable sets. Clearly, $M\subset \tilde{M}$, if $M = \tilde{M}$, then $\mu$ is saturated.

a.) If $\mu$ is $\sigma$-finite, then $\mu$ is saturated.

Proof - Suppose $\mu$ is $\sigma$-finite. Let $A\in\tilde{M}$, and let $X = \bigcup_{1}^{\infty}E_j$ where $E_j\in M$ and $\mu(E_j) < \infty$ for all $j$. We know $M\subset \tilde{M}$ what we want to show is that $\tilde{M}\subset M$. We can write $$A = A\cap X = A\cap \left(\bigcup_{1}^{\infty}E_j\right) = \bigcup_{1}^{\infty}E_j\cap A$$ Since $\mu(E_j)<\infty$ we have $E_j\cap A\in M$ for all $j$. Therefore, $\tilde{M}\subset M$, thus $\mu$ is saturated.

b.) $\tilde{M}$ is $\sigma$-algebra.

Proof -

i.) $\emptyset\in M\subset \tilde{M}$, so $\emptyset\in \tilde{M}$.

ii.) Let $B\in \tilde{M}$. Take any $E\in M$ such that $\mu(E) < \infty$. Then $$E\setminus B = E\cap B^c = E\cap (E\cap B)^c$$ since $E\in M$ and $(E\cap B)\in M$ then $(E\cap (E\cap B)^c\in M$. Thus we have $B^c\in \tilde{M}$.

iii.) Let $\{B_j\}_{1}^{\infty}\in \tilde{M}$. Take any $E\in M$ with $\mu(E)< \infty$. Then, $$\left(\bigcup_{1}^{\infty}B_j\right)\cap E = \bigcup_{1}^{\infty}(B_j\cap E)\in M$$ so, by definition of $\tilde{M}$, $\bigcup_{1}^{\infty}B_j\in \tilde{M}$. Therefore $\tilde{M}$ us a $\sigma$-algebra.

c.) Define $\tilde{\mu}$ on $\tilde{M}$ by $\tilde{\mu}(E) = \mu(E)$ if $E\in M$ and $\tilde{\mu}(E) = \infty$ otherwise. Then $\tilde{\mu}$ is a saturated measure on $\tilde{M}$, called the saturation of $\mu$.

Step 1: Show that $\tilde{\mu}$ is a measure on $\tilde{M}$.

Proof -

i.) $\tilde{\mu}(\emptyset) = \mu(\emptyset) = 0$.

ii.) Let $\{E_j\}_{1}^{\infty}\in \tilde{M}$ that is pairwise disjoint. Let $$E = \bigcup_{1}^{\infty}E_j$$ If $E \in M$ then \begin{align*} \tilde{\mu}(E) = \tilde{\mu}\left(\bigcup_{1}^{\infty}E_j\right) &= \mu\left(\bigcup_{1}^{\infty}E_j\right)\\ &= \sum_{1}^{\infty}\mu(E_j)\\ &= \sum_{1}^{\infty}\tilde{\mu}(E_j) \end{align*} If $E\notin M$ then $\tilde{\mu}(E) = \infty$... not sure where to go from here.

Step 2 - $\tilde{\mu}$ is saturated.

Proof - Let $E\subset X$ such that $E\cap A\in \tilde{M}$ when $\tilde{\mu}(A) < \infty$. Choose a $B\in M$ such that $\mu(B)<\infty$. Then, clearly $\tilde{\mu}(B)<\infty$ and $E\cap B\in \tilde{M}$. So, $E\cap B = (E\cap B)\cap B\in M$ so $E\in \tilde{M}$ it thus follows that $\tilde{\mu}$ is saturated.

d.) If $\mu$ is complete, so is $\tilde{\mu}$.

Proof - Suppose $\mu$ is complete. Let $A\subset X$ and suppose there is a $B\in \tilde{M}$ such that $A\subset B$ and $\mu(B) = 0$. Since $B\in\tilde{M}$ and $\mu(B) = 0$ then $\tilde{\mu}(B) < \infty$ and hence $B\in M$. This, since $A\subset B$ we have $A\in M$ by completeness of $\mu$. Therefore, $A\in \tilde{M}$ and $\tilde{\mu}$ is complete.

e.) Suppose that $\mu$ is semifinite. For $E\in\tilde{M}$, define $\underline{\mu}(E) = \sup\{\mu(A):A\in M, A\subset E\}$. Then $\underline{\mu}$ is a saturated measure on $\tilde{M}$ that extends $\mu$.

Step 1 - $\underline{\mu}$ is a measure.

Proof -

i.) $\overline{\mu}(\emptyset) = \mu(\emptyset) = 0$

ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $\tilde{M}$. Set, $$E = \bigcup_{1}^{\infty}E_j$$ then by definition of $\tilde{M}$ there is an $A\in M$ and $A\subset E$.

Case 1 - $\mu(A) < \infty$. Then $$\mu(A) = \mu\left(\bigcup_{1}^{\infty}E_j\cap A\right) = \sum_{1}^{\infty}\mu(E_j\cap A)\leq \sum_{1}^{\infty}\underline{\mu}(E_j)$$

Case 2 - $\mu(A) = \infty$. By semifiniteness, for all $C>0$ there exists a $F\subset A$ such that $F\in M$ and $\mu(F) = C$. Then by case 1, $\leq \sum_{1}^{\infty}\underline{\mu}(E_j) = \infty$. Therefore, $\mu(A) \leq \sum_{1}^{\infty}\underline{\mu}(E_j)$. Taking the supremum over all $A$ we have $$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)\leq \sum_{1}^{\infty}\underline{\mu}(E_j)$$ Now we need to show the reverse inequality. By the definition of supremum there exists a sequence $\{B_i\}_{1}^{\infty}\in M$ and $B_i\subset E_i$ for all $i$. Thus, $\underline{\mu}(E_i)\leq \mu(B_i) + \epsilon 2^{-i}$. Therefore, \begin{align*} \sum_{1}^{\infty}\underline{\mu}(E_i) &\leq \sum_{1}^{\infty}\mu(B_i) + \epsilon\\ &= \mu\left(\bigcup_{1}^{\infty}B_i\right) + \epsilon \ \ \text{is this true because of case 1?}\\ &\leq \underline{\mu}\left(\bigcup_{1}^{\infty}E_i\right) + \epsilon \end{align*} Since this holds for all $\epsilon > 0$, we have $$\sum_{1}^{\infty}\underline{\mu}(E_j)\leq \underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)$$ Therefore, $$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right) = \sum_{1}^{\infty}\underline{\mu}(E_j)$$ and hence $\underline{\mu}$ is a measure.

Step 2 - $\underline{\mu}$ is saturated.

Proof - Let $E\subset X$ be such that $E\cap A\in \tilde{M}$ when $\underline{\mu}(A)< \infty$. Take any $B\in M$ such that $\mu(B) < \infty$. Then $\underline{\mu}(B) < \infty$ so $E\cap B\in \tilde{M}$. Thus, $E\cap B = (E\cap B)\cap B\in M$ and $E\in\tilde{M}$, hence, $\underline{\mu}$ is saturated.

Step 3 - $\underline{\mu}$ is an extention of $\mu$.

Proof - Let $E\in M$. For any $A\in M$ such that $A\subset E$, we have by monotonicity that $\mu(A)\leq \mu(E)$. Since $\underline{\mu}(E)$ is the supremum over all such $A$, we must have that $\underline{\mu}(E)\leq \mu(E)$. OTOH.... not sure how to show the reverse inequality.

f.) Let $X_1$ and $X_2$ be disjoint uncountable sets, $X = X_1\cup X_2$, and $M$ the $\sigma$-alegbra of countable or co-countable sets in $X$. Let $\mu_0$ be counting measure on $\mathcal{P}(X_1)$ and define $\mu$ on $M$ by $\mu(E) = \mu_0(E\cap X_1)$. Then $\mu$ is a measure on $M$, $\tilde{M} = \mathcal{P}(X)$, and in the notation of parts (c) and (e), $\tilde{\mu}\neq \underline{\mu}$.

Step 1 - $\mu$ is a measure on $M$.

Proof -

i.) $\mu(\emptyset) = \mu_0(\emptyset\cap X_1) = 0$

ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$, then \begin{align*} \mu\left(\bigcup_{1}^{\infty}E_j\right) &= \mu_0\left(\bigcup_{1}^{\infty}E_j\cap X_1\right)\\ &= \sum_{1}^{\infty}\mu_0(E_j\cap X_1)\\ &= \sum_{1}^{\infty}\mu(E_j) \ \ \ \text{is this true because} \ \mu_0 \ \text{is a counting measure?} \end{align*} Therefore $\mu$ is a measure on $M$.

Step 2 - $\tilde{M} = \mathcal{P}(X)$

Proof -

Step 3 - $\tilde{\mu}\neq \underline{\mu}$

Proof - Take $y_1,y_2\in X_1$. Let $E = \{y_1,y_2\}\cup X_2$. Then $E\notin M$, so $\tilde{\mu}(E) = \infty$. However, $\underline{\mu}(E) = 2$.

I will re-edit my question and include these other proofs as I continue to do them. I am pretty sure my proof for $\tilde{\mu}$ is a measure is incorrect. But I am not really sure how to do it. Any suggestions on any of these is greatly appreciated.

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  • $\begingroup$ a) I would remove ' Also since A is locally measurable and is a countable union of these sets, A∈M' and just say "hence $A\in M$. $\endgroup$ – YannickSSE May 31 '16 at 11:19
  • $\begingroup$ In c) Step 2, $E\cap B\cap A$ does not make sense to me. You can end that part by using the following: as $E\cap B\in \tilde M$, $E\cap B=(E\cap B)\cap B\in M$. So, $E\in \tilde M$. $\endgroup$ – Emre May 31 '16 at 11:19
  • $\begingroup$ Also, there are tons of typos. $\endgroup$ – Emre May 31 '16 at 11:21
  • $\begingroup$ I see thank you for the suggestions I will edit those parts. Although still not sure how to show $\tilde{\mu}$ is a measure. $\endgroup$ – Wolfy May 31 '16 at 11:21
  • $\begingroup$ @Wolfy non-negativity, well-definedness etc. are obvious. So, it is enough to show the additivity. For that part, you can use my hint in the deleted comment. Let me elaborate a little more: if $E_k\in M$ for all $k$, we have the additivity by the fact that $\mu$ is a measure. So, suppose $E_k\notin M$ for some $k$. Then, $\tilde\mu(E_k)=\infty$. So, the sum on the LHS is $\infty$. So, it is sufficient to prove that $\tilde\mu(\bigcup_kE_k)=\infty$. Suppose the contrary. Then, $E=\bigcup_kE_k\in M$ and $\mu(E)<\infty$. Then, use this to get a contradiction. $\endgroup$ – Emre May 31 '16 at 11:29
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Items a and b are correct. For the rest of the items, some need just minor improvements and some really need to be fixed. I tried to keep the proof as close to your proof as possible.

c.) Define $\tilde{\mu}$ on $\tilde{M}$ by $\tilde{\mu}(E) = \mu(E)$ if $E\in M$ and $\tilde{\mu}(E) = \infty$ otherwise. Then $\tilde{\mu}$ is a saturated measure on $\tilde{M}$, called the saturation of $\mu$.

Step 1: Show that $\tilde{\mu}$ is a measure on $\tilde{M}$.

Proof -

It is clear that $\tilde{\mu}$ is a non-negative well-defined function on $\tilde{M}$.

i.) $\tilde{\mu}(\emptyset) = \mu(\emptyset) = 0$.

ii.) Let $\{E_j\}_{1}^{\infty}\in \tilde{M}$ that is pairwise disjoint. Let $$E = \bigcup_{1}^{\infty}E_j$$ If $E \in M$ then \begin{align*} \tilde{\mu}(E) = \tilde{\mu}\left(\bigcup_{1}^{\infty}E_j\right) &= \mu\left(\bigcup_{1}^{\infty}E_j\right) = \sum_{1}^{\infty}\mu(E_j)= \sum_{1}^{\infty}\tilde{\mu}(E_j) \end{align*} If $E\notin M$ then $\tilde{\mu}(E) = \infty$. Since $E = \bigcup_{1}^{\infty}E_j$ and $M$ is a $\sigma$-algebra, there is $j_0$ such that $E_{j_0}\notin M$. So $\tilde{\mu}(E_{j_0}) = \infty$. So we have $$\tilde{\mu}(E) = \infty = \tilde{\mu}(E_{j_0})\leqslant \sum_{1}^{\infty}\tilde{\mu}(E_j)\leqslant \infty$$ So we have $$\tilde{\mu}(E) = \infty = \sum_{1}^{\infty}\tilde{\mu}(E_j)$$

Step 2 - $\tilde{\mu}$ is saturated.

Proof - Let $E\subset X$ such that $E\cap A\in \tilde{M}$ when $\tilde{\mu}(A) < \infty$. For any $B\in M$ such that $\mu(B)<\infty$, we clearly have $\tilde{\mu}(B)=\mu(B)<\infty$ and $E\cap B\in \tilde{M}$. Now, since $E\cap B\in \tilde{M}$ and $B\in M$ such that $\mu(B)<\infty$ we have $E\cap B = (E\cap B)\cap B\in M$. So we proved that, for any $B\in M$ such that $\mu(B)<\infty$, $E\cap B \in M$. So $E\in \tilde{M}$. It thus follows that $\tilde{\mu}$ is saturated.

d.) If $\mu$ is complete, so is $\tilde{\mu}$.

Proof - Suppose $\mu$ is complete. Let $A\subset X$ and suppose there is a $B\in \tilde{M}$ such that $A\subset B$ and $\tilde{\mu}(B) = 0$. Since $B\in \tilde{M}$ and $\tilde{\mu}(B) = 0 < \infty$ and hence $ B \in M$ and $\mu(B)=\tilde{\mu}(B) = 0$. Since $A\subset B$ we have $A\in M \subset \tilde{M}$ by completeness of $\mu$. Therefore, $A\in \tilde{M}$ and $\tilde{\mu}$ is complete.

e.) Suppose that $\mu$ is semifinite. For $E\in\tilde{M}$, define $\underline{\mu}(E) = \sup\{\mu(A):A\in M, A\subset E\}$. Then $\underline{\mu}$ is a saturated measure on $\tilde{M}$ that extends $\mu$.

Before proving the item e we prove a lemma

Lemma: Suppose that $\mu$ is semifinite. For $E\in\tilde{M}$, $$\underline{\mu}(E) = \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$

Proof of the lemma:

We clearly have
$$\underline{\mu}(E)=\sup\{\mu(A):A\in M, A\subset E \} \geqslant \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$

Case 1: If, for all $A\in M$, $A\subset E$, we have $\mu(A)<\infty$ then we have $$\underline{\mu}(E)=\sup\{\mu(A):A\in M, A\subset E \} = \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$

Case 2: Now, suppose there is $A_0\in M$, $A_0\subset E$ such that $\mu(A_0)=\infty$. Then $\underline{\mu}(E)=\infty$ and, since $\mu$ is semifinite, we have that $$\sup\{\mu(A):A\in M, A\subset A_0 \textrm{ and } \mu(A)<\infty\}=\mu(A_0)=\infty$$ So we have $$\infty=\underline{\mu}(E)=\sup\{\mu(A):A\in M, A\subset E \} \geqslant \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}\geqslant \\ \geqslant \sup\{\mu(A):A\in M, A\subset A_0 \textrm{ and } \mu(A)<\infty\}=\infty$$ So we have $$\underline{\mu}(E)=\infty= \sup\{\mu(A):A\in M, A\subset E \textrm{ and } \mu(A)<\infty\}$$

End of proof of the lemma

Step 1 - $\underline{\mu}$ is a measure.

Proof -

It is clear that $\tilde{\mu}$ is a non-negative well-defined function on $\tilde{M}$.

i.) $\underline{\mu}(\emptyset) = \mu(\emptyset) = 0$

ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $\tilde{M}$. Set, $$E = \bigcup_{1}^{\infty}E_j$$ Then, by the lemma, we have: \begin{align*} \underline{\mu}\left ( \bigcup_{1}^{\infty}E_j \right ) & = \sup\left\{\mu(A):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\} \end{align*} Since each $E_j \in \tilde{M}$, then, for each $A\in M$, $A\subset \bigcup_{1}^{\infty}E_j$ and $\mu(A)<\infty$, we have $E_j \cap A \in M$ and $$\mu(A)=\sum_1^\infty\mu(E_j \cap A)$$ So we have \begin{align*} \underline{\mu}\left ( \bigcup_{1}^{\infty}E_j \right ) & = \sup\left\{\mu(A):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\}=\\ & = \sup\left\{\sum_1^\infty\mu(A \cap E_j):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\}\leqslant \\ &\leqslant \sum_1^\infty\sup\left\{\mu(A \cap E_j):A\in M, A\subset \bigcup_{1}^{\infty}E_j \textrm{ and } \mu(A)<\infty \right\}\leqslant \\ &\leqslant\sum_1^\infty\sup\left\{\mu(B):B\in M, B\subset E_j \textrm{ and } \mu(B)<\infty \right\}= \\ &= \sum_1^\infty \underline{\mu}( E_j ) & \leqslant \end{align*}

So we proved $$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)\leqslant \sum_{1}^{\infty}\underline{\mu}(E_j)$$

Now note that from the defintion of $\overline{\mu}$ we have \begin{align*} \sum_{1}^{\infty}\underline{\mu}(E_j)&= \sum_{1}^{\infty}\sup\left\{\mu(B_j):B_j\in M, B_j\subset E_j \right\}= \\ &= \sup\left\{\sum_{1}^{\infty}\mu(B_j):B_j\in M, B_j\subset E_j \right\}= \\ &= \sup\left\{\mu(\bigcup_{1}^{\infty}B_j):B_j\in M, B_j\subset E_j \right\}\leqslant \\ &\leqslant \sup\left\{\mu(B):B\in M, B\subset \bigcup_{1}^{\infty} E_j \right\}= \\ &=\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right) \end{align*} So we proved $$\underline{\mu}\left(\bigcup_{1}^{\infty}E_j\right)= \sum_{1}^{\infty}\underline{\mu}(E_j)$$ and hence $\underline{\mu}$ is a measure.

Step 2 - $\underline{\mu}$ is saturated.

Proof - Let $E\subset X$ be such that $E\cap A\in \tilde{M}$ when $A \in \tilde{M}$ and $\underline{\mu}(A)< \infty$.

Take any $B\in M$ such that $\mu(B) < \infty$. Then $\underline{\mu}(B) \leqslant \mu(B) < \infty$ so $E\cap B\in \tilde{M}$. Thus, $E\cap B = (E\cap B)\cap B\in M$. So, we proved that, for any $B\in M$ such that $\mu(B) < \infty$, $E\cap B \in M$. So $E\in\tilde{M}$, hence, $\underline{\mu}$ is saturated.

Step 3 - $\underline{\mu}$ is an extention of $\mu$.

Proof - Let $E\in M$. Then $$\underline{\mu}(E)=\sup\{\mu(A): A\in M, A\subset E \}=\mu(E)$$ So $\underline{\mu}$ is an extention of $\mu$.

f.) Let $X_1$ and $X_2$ be disjoint uncountable sets, $X = X_1\cup X_2$, and $M$ the $\sigma$-alegbra of countable or co-countable sets in $X$. Let $\mu_0$ be counting measure on $\mathcal{P}(X_1)$ and define $\mu$ on $M$ by $\mu(E) = \mu_0(E\cap X_1)$. Then $\mu$ is a measure on $M$, $\tilde{M} = \mathcal{P}(X)$, and in the notation of parts (c) and (e), $\tilde{\mu}\neq \underline{\mu}$.

Step 1 - $\mu$ is a measure on $M$.

Proof -

i.) $\mu(\emptyset) = \mu_0(\emptyset\cap X_1) = 0$

ii.) Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$, then \begin{align*} \mu\left(\bigcup_{1}^{\infty}E_j\right) &= \mu_0\left(\bigcup_{1}^{\infty}E_j\cap X_1\right)\\ &= \sum_{1}^{\infty}\mu_0(E_j\cap X_1)\\ &= \sum_{1}^{\infty}\mu(E_j) \ \ \ \text{is this true because} \ \mu_0 \ \text{is a counting measure?} \end{align*} Therefore $\mu$ is a measure on $M$.

Step 2 - $\tilde{M} = \mathcal{P}(X)$

Proof - Note that if $A \in M$ and $\mu(A)<\infty$ then $A \cap X_1$ is a finite set, so $A$ can not be co-countable. So $A$ is countable.

Given any set $E\subset X$, then for all $A \in M$ and $\mu(A)<\infty$, $A$ is countable and so is $E \cap A$. So $E \cap A \in M$. So $\tilde{M}=\mathcal{P}(X)$.

Step 3 - $\tilde{\mu}\neq \underline{\mu}$

Proof - Take $y_1,y_2\in X_1$. Let $E = \{y_1,y_2\}\cup X_2$. Then $E\notin M$, so $\tilde{\mu}(E) = \infty$. However, $\underline{\mu}(E) = 2$.

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  • $\begingroup$ 16.(f)step 2 why can't A be co-countable? $\endgroup$ – ZHU Sep 30 '16 at 1:43
  • $\begingroup$ @ZHU, Since $X_1$ is uncountable and $A\cap X_1$ is finite, we have that $X_1\setminus A=X_1\setminus (A\cap X_1)$ is uncountable. Since $X \setminus A \supset X_1\setminus A$, we have that $X\setminus A$ is uncountable. So $A$ is not co-countable. $\endgroup$ – Ramiro Sep 30 '16 at 3:51

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